Integral Sum Problem #272: Find the Integral Part of a Sum

[anemone]

Here is this week's POTW:

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Find the integral part of $$\large \sum_{n=1}^{10^9} n^{\small-\dfrac{2}{3}}$$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Opalg for his correct solution, which you can find below::)

Spoiler

For convenience, write $N$ for $10^9$.

Since $x^{-2/3}$ is a decreasing function (for $x>0$), and has integral $3x^{1/3}$, we can use approximations by Riemann sums on the intervals $[2,N+1]$ and $[1,N]$ to get $$\int_2^{N+1}x^{-2/3}dx < \sum_2^Nn^{-2/3} < \int_1^Nx^{-2/3}dx,$$ or $$3\bigl((N+1)^{1/3} - 2^{1/3} \bigr) < \sum_2^Nn^{-2/3} < 3(N^{1/3}-1).$$ But $N^{1/3} = 1000$ and $(N+1)^{1/3} > 1000$, so we get $$2996.22 < \sum_2^Nn^{-2/3} < 2997.$$ Now add the first term of the sum (which is $1$), to get $$2997 < \sum_1^Nn^{-2/3} < 2998.$$ So the integer part of $$\sum_1^Nn^{-2/3}$$ is $2997.$